Problem: You have found the following ages (in years) of all 4 bears at your local zoo: $ 10,\enspace 14,\enspace 18,\enspace 15$ What is the average age of the bears at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 4 bears at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{10 + 14 + 18 + 15}{{4}} = {14.3\text{ years old}} $ Find the squared deviations from the mean for each bear. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $10$ years $-4.3$ years $18.49$ years $^2$ $14$ years $-0.3$ years $0.09$ years $^2$ $18$ years $3.7$ years $13.69$ years $^2$ $15$ years $0.7$ years $0.49$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{18.49} + {0.09} + {13.69} + {0.49}} {{4}} $ $ {\sigma^2} = \dfrac{{32.76}}{{4}} = {8.19\text{ years}^2} $ The average bear at the zoo is 14.3 years old. The population variance is 8.19 years $^2$.